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It is known that in △ ABC, D and E are points on BC and ab respectively. AD and CE intersect F and CD = 1 3BC，AE=2 5ab. Find s △ ACF The value of s △ CDF
Then egbg = CDBD, ∵ CD = 13bc, ᙽ BD = 2dc, ∵ BG = 2EG, ? AE = 25ab, AE: be = 2:3, ? AE = 2EG, ? CE ∥ DG, ? AFFD = AEEG = 2eeg = 2, ? AFC edge AF is equal to the height of △ CDF edge DF. Let this height be h, s ﹤ ACFs ﹤ CDF
As shown in the figure, in △ ABC, ∠ ACB = 90 °, ad ⊥ AB, ad = AB, be ⊥ DC in E, AF ⊥ AC crossing EB in F, verification: ﹣ ACF = ∠ BCF Help The picture is in my space
prove:
Extend AB, de so that they intersect at a point G,
Then the angle ADG = angle GBE = angle ABF is known by ad ⊥ AB, be ⊥ DC,
Because angle CAF = angle bad
Then the angle CAF angle BAC = angle bad angle BAC, namely angle Fab = angle DAC,
And because ad = AB,
So from AAS, we know that the triangle BAF is equal to the triangle DAC
If AF = AC, then angle AFC = angle ACF,
Because AF is parallel to BC, angle AFC= angle BFC,
So angle ACF = angle BFC
As shown in the figure, in △ ABC, point E is on AB, point D is on BC, BD = be, ∠ bad = ∠ BCE, ad and CE intersect at point F. try to judge the shape of △ AFC and explain the reasons
The △ AFC is an isosceles triangle
In △ bad and △ BCE,
B = ∠ B (common angle), ∠ bad = ∠ BCE, BD = be,
∴△BAD≌△BCE（AAS），
∴BA=BC，∠BAD=∠BCE，
∴∠BAC=∠BCA，
Ψ BAC - ∠ bad = ∠ BCA - ∠ BCE, i.e., ∠ fac = ∠ FCA
∴AF=CF，
The △ AFC is an isosceles triangle
As shown in the figure, in the isosceles trapezoid ABCD, ab ∥ CD, extend the bottom edge AB to e so that be = DC Confirmation: AC = CE
Proof: connect BD
∵AB∥CD，BE=DC，
The quadrilateral becd is a parallelogram,
∴CE=BD，
∵ quadrilateral ABCD is isosceles trapezoid,
∴AC=BD，
∴AC=CE．
As shown in the figure, in the isosceles trapezoid ABCD, known ad ∥ BC, ab = DC, ad = 2, BC = 4, extend BC to e, so that CE = ad (1) Write out all triangles which are congruent with △ DCE, and choose one pair to explain the reason of congruence; (2) When the high DF of isosceles trapezoid ABCD is, the diagonal AC and BD are perpendicular to each other? Please answer and explain why
(1) Δ CDA ≌ △ DCE, △ bad ≌ △ DCE; (2 points)
① The reasons for △ CDA ≌ △ DCE are as follows:
∵AD∥BC，
Ψ CDA = ∠ DCE. (3 points)
And ∵ Da = CE, CD = DC, (4 points)
≌ △ CDA ≌ △ DCE. (5 points)
② The reasons for △ bad ≌ △ DCE are as follows:
∵AD∥BC，
Ψ CDA = ∠ DCE. (3 points)
The quadrilateral ABCD is isosceles trapezoid,
∴∠BAD=∠CDA，
Ψ bad = ∠ DCE. (4 points)
AB = CD, ad = CE,
≌△ DCE. (5 points)
(2) When the height DF of the isosceles trapezoid ABCD is 3, the diagonal AC and BD are perpendicular to each other. (6 points)
The reason is: let the intersection of AC and BD be point G, ∵ quadrilateral ABCD is isosceles trapezoid,
∴AC=DB．
And ∵ ad = CE, ad ∵ BC,
The quadrilateral aced is a parallelogram, (7 points)
∴AC=DE，AC∥DE．
ν DB = de. (8 points)
Then BF = Fe,
And ∵ be = BC + CE = BC + ad = 4 + 2 = 6,
ν BF = Fe = 3. (9 points)
∵DF=3，
∴∠BDF=∠DBF=45°，∠EDF=∠DEF=45°，
∴∠BDE=∠BDF+∠EDF=90°，
And ∵ AC ∥ De
Ψ BGC = ∠ BDE = 90 ° i.e. AC ⊥ BD. (10 points)
(Note: if DF = BF = Fe, get ∠ BDE = 90 ° and give full marks as well.)
As shown in the figure, in the isosceles trapezoid ABCD, known ad ∥ BC, ab = DC, ad = 2, BC = 4, extend BC to e, so that CE = ad (1) Write out all triangles which are congruent with △ DCE, and choose one pair to explain the reason of congruence; (2) When the high DF of isosceles trapezoid ABCD is, the diagonal AC and BD are perpendicular to each other? Please answer and explain why
(1) Δ CDA ≌ △ DCE, △ bad ≌ △ DCE; (2 points)
① The reasons for △ CDA ≌ △ DCE are as follows:
∵AD∥BC，
Ψ CDA = ∠ DCE. (3 points)
And ∵ Da = CE, CD = DC, (4 points)
≌ △ CDA ≌ △ DCE. (5 points)
② The reasons for △ bad ≌ △ DCE are as follows:
∵AD∥BC，
Ψ CDA = ∠ DCE. (3 points)
The quadrilateral ABCD is isosceles trapezoid,
∴∠BAD=∠CDA，
Ψ bad = ∠ DCE. (4 points)
AB = CD, ad = CE,
≌△ DCE. (5 points)
(2) When the height DF of the isosceles trapezoid ABCD is 3, the diagonal AC and BD are perpendicular to each other. (6 points)
The reason is: let the intersection of AC and BD be point G, ∵ quadrilateral ABCD is isosceles trapezoid,
∴AC=DB．
And ∵ ad = CE, ad ∵ BC,
The quadrilateral aced is a parallelogram, (7 points)
∴AC=DE，AC∥DE．
ν DB = de. (8 points)
Then BF = Fe,
And ∵ be = BC + CE = BC + ad = 4 + 2 = 6,
ν BF = Fe = 3. (9 points)
∵DF=3，
∴∠BDF=∠DBF=45°，∠EDF=∠DEF=45°，
∴∠BDE=∠BDF+∠EDF=90°，
And ∵ AC ∥ De
Ψ BGC = ∠ BDE = 90 ° i.e. AC ⊥ BD. (10 points)
(Note: if DF = BF = Fe, get ∠ BDE = 90 ° and give full marks as well.)
As shown in the figure, in the isosceles trapezoid ABCD, known ad ∥ BC, ab = DC, ad = 2, BC = 4, extend BC to e, so that CE = ad (1) It is proved that △ bad ≌ △ DCE; (2) If AC ⊥ BD, find the value of high DF of isosceles trapezoid ABCD
(1) Proof: ∵ ad ∥ BC,
Ψ CDA = ∠ DCE. (1 point)
The quadrilateral ABCD is isosceles trapezoid,
ν CDA = (bad)
Ψ bad = ∠ DCE. (3 points)
∵AB=DC，AD=CE，
≌△ DCE; (5 points)
（2）∵AD=CE，AD∥BC，
The quadrilateral aced is a parallelogram, (7 points)
/ / AC ∥ de. (8 points)
∵AC⊥BD，
⊥ BD. (9 points)
It can be seen from (1) that △ bad ≌ △ DCE,
ν de = BD. (10 points)
Therefore, △ BDE is an isosceles right triangle, that is ∠ e = 45 °,
/ / DF = Fe = FC + CE. (12 points)
∵ the quadrilateral ABCD is isosceles trapezoid, while ad = 2, BC = 4,
∴FC=1
2（BC-AD）=1
2 (4-2) = 1. (13 points)
∵CE=AD=2，
/ / DF = 3. (14 points)
In the known parallelogram ABCD, e is the midpoint of AD and the extension line of CE intersects BA at point F Verification: CD = AF 2. If BC = 2CD, it is proved that: ∠ f = ∠ BCF Why no one answers
1. proof: in △ FAE and △ CDE
Because ∠ f = ECD
∠FEA＝∠CED
AE＝ED
So △ FAE ≌ △ CDE
So CD = AF
2. Proof: because CD = AF
AB＝CD
So AF + AB = 2CD
So BF = BC
So ∠ f = ∠ BCF
In the parallelogram ABCD, e is the midpoint of AD, and the extension line of CE intersects BA at point F If BC = 2CD, find: ∠ f = ∠ BCF
prove:
∵AB‖CD
∴∠F=∠ECD,∠FAE=∠D
∵AE=DE
∴△AEF≌△DEC
∴AF=CD
∵ quadrilateral ABCD is a parallelogram
∴AB=CD,AD=BC
∴BF=2CD
∴BF=BC
∴∠F=∠BCF
As shown in the figure, in the rectangle ABCD, ab = 2, BC = 4, and the vertical bisector of diagonal AC is called AD and BC respectively at points E and F, connecting CE, then what is the length of CE As shown in the figure, in the rectangle ABCD, ab = 2, BC = 3, and the vertical bisectors of diagonal AC are called AD and BC respectively at points E and F, connecting CE. What is the length of CE
Solution: let the length of EC be X,
Because EF is the vertical bisector of AC, so △ AEO and △ eco are congruent, so AE = EC, ed = ad-ae = 4-x
According to Pythagorean theorem, EC ^ 2 = ed ^ 2 + DC ^ 2 in RT △ EDC
X^2=(4-X)^2+2^2
X^2=4^2-2*4*X+X^2+2^2
X^2=16-8X+4+X^2
8X=20
X=5/2
A: the length of CE is 5 / 2
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